What is the slope of the line tangent to $f(x) = x^{2}-2x-6$ at $x = 0$ ?
Explanation: The slope of the tangent line is $ \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$ $ = \lim_{h \to 0} \frac{((x+h)^{2}-2(x+h)-6) - (x^{2}-2x-6)}{h}$ $ = \lim_{h \to 0} \frac{(x^{2}+2x h+h^{2}-2(x+h)-6) - (x^{2}-2x-6)}{h}$ $ = \lim_{h \to 0} \frac{x^{2}+2(x h)+h^{2}-2x-2h-6-x^{2}+2x+6}{h}$ $ = \lim_{h \to 0} \frac{2(x h)+h^{2}-2h}{h}$ $ = \lim_{h \to 0} 2x+h-2$ $ = 2x-2$ $ = (2)(0)-2$ $ = -2$